Question

Solve the following:
\[\int {\dfrac{2}{{(1 - x)(1 + {x^2})}}dx} \]

Answer 1

Hint: The given problem is an indefinite integral since there are no limit points. We will first decompose the function inside the integral. Then we will integrate the parts one by one. Decomposition of the function helps us to solve the problem easily.
Formula: Differentiation formula that we need to know is \[{x^n}dx = n{x^{n - 1}}\]. And the integration formulas that we need to know are \[\int {cf(x)dx} = c\int {f(x)dx} \] where \[c\] is a constant, and \[\int {\dfrac{1}{x}dx = \ln (x)} \].

Complete step by step answer:
It is given that \[\int {\dfrac{2}{{(1 - x)(1 + {x^2})}}dx} \].
Using the formula \[\int {cf(x)dx} = c\int {f(x)dx} \] let us rewrite the given function.
\[\int {\dfrac{2}{{(1 - x)(1 + {x^2})}}dx} = 2\int {\dfrac{1}{{(1 - x)(1 + {x^2})}}dx} \]
Let \[f(x) = \dfrac{1}{{(1 - x)(1 + {x^2})}}\]
Let us first decompose \[f(x)\] using partial fraction decomposition.
On decomposing we get \[f(x) = \dfrac{{x + 1}}{{2({x^2} + 1)}} - \dfrac{1}{{2(x - 1)}}\]
Let us substitute it in the integral function.
\[2\int {\dfrac{1}{{(1 - x)(1 + {x^2})}}dx} = 2\int {\dfrac{{x + 1}}{{2({x^2} + 1)}} - \dfrac{1}{{2(x - 1)}}dx} \]
Let us now split the integral.
\[ = 2\int {\dfrac{{x + 1}}{{2({x^2} + 1)}}dx - 2\int {\dfrac{1}{{2(x - 1)}}dx} } \]
Using the formula \[\int {cf(x)dx} = c\int {f(x)dx} \] again.
\[ = \int {\dfrac{{x + 1}}{{({x^2} + 1)}}dx - \int {\dfrac{1}{{(x - 1)}}dx} } \]
Now let us start to integrate the second term.
Let \[u = x - 1\]. On differentiating this concern, the variable \[x\] we get \[du = dx\] let us substitute them in the above expression.
 \[ = \int {\dfrac{{x + 1}}{{{x^2} + 1}}dx - \int {\dfrac{1}{u}du} } \]
Using the formula \[\int {\dfrac{1}{x}dx = \ln (x)} \] we get
 \[ = \int {\dfrac{{x + 1}}{{{x^2} + 1}}dx - \ln (u)} \]
Let’s re-substitute \[u = x - 1\].
 \[ = \int {\dfrac{{x + 1}}{{{x^2} + 1}}dx - \ln (x - 1)} \]
Now let us integrate the first function. Let’s split that fraction first.
\[ = \int {\left( {\dfrac{x}{{{x^2} + 1}} + \dfrac{1}{{{x^2} + 1}}} \right)dx - \ln (x - 1)} \]
Now let us split the integral.
\[ = \int {\dfrac{x}{{{x^2} + 1}}dx + \int {\dfrac{1}{{{x^2} + 1}}} dx - \ln (x - 1)} \]
Now let us substitute\[u = {x^2} + 1\]. On differentiating this concern, the variable \[x\]we get \[xdx = \dfrac{{du}}{2}\]let us substitute them in the above expression.
\[ = \int {\dfrac{1}{{2u}}du + \int {\dfrac{1}{{{x^2} + 1}}} dx - \ln (x - 1)} \]
Again, using the formula \[\int {cf(x)dx} = c\int {f(x)dx} \] we get
\[ = \dfrac{1}{2}\int {\dfrac{1}{u}du + \int {\dfrac{1}{{{x^2} + 1}}} dx - \ln (x - 1)} \]
Now let’s integrate it using the formula \[\int {\dfrac{1}{x}dx = \ln (x)} \].
\[ = \dfrac{{\ln (u)}}{2} + \int {\dfrac{1}{{{x^2} + 1}}} dx - \ln (x - 1)\]
Now let’s re-substitute \[u = {x^2} + 1\].
\[ = \dfrac{{\ln ({x^2} + 1)}}{2} + \int {\dfrac{1}{{{x^2} + 1}}} dx - \ln (x - 1)\]
Now let’s integrate the second part. Using the formula \[\int {\dfrac{1}{{{x^2} + 1}}dx = {{\tan }^{ - 1}}(x)} \]
\[ = \dfrac{{\ln ({x^2} + 1)}}{2} + {\tan ^{ - 1}}(x) - \ln (x - 1) + C\], where \[C\] is the integration constant
Thus, \[\int {\dfrac{2}{{(1 - x)(1 + {x^2})}}dx} = \dfrac{{\ln ({x^2} + 1)}}{2} + {\tan ^{ - 1}}(x) - \ln (x - 1) + C\]

Note: There are two types of integral available, they are definite and indefinite. Definite integral will have limit points and indefinite integrals won’t have any limit points. If there is a limit point, we need to substitute it after the integration.